Movie Hook: Tuning the Radio in Vintage Films
In old movies, you see characters turning a dial to tune a radio to different stations. That dial adjusts a variable capacitor in an LCR circuit. When the circuit’s resonant frequency matches the radio wave frequency, the circuit “lights up” with maximum current - you hear that station loud and clear! Every frequency has its own LCR fingerprint. The same principle works in your phone’s antenna, Wi-Fi receivers, and even MRI machines. Welcome to the world of resonance!
What is an LCR Circuit?
An LCR circuit contains a resistor (R), inductor (L), and capacitor (C) connected to an AC source.
LCR Circuit: Combines resistive, inductive, and capacitive elements.
Key phenomenon: Resonance - maximum current at specific frequency
Applications:
- Radio/TV tuning
- Filters (pass or block specific frequencies)
- Oscillators
- Impedance matching
Series LCR Circuit
Setup: R, L, and C connected in series with AC source $V = V_0\sin(\omega t)$
Same current flows through all elements (series connection).
Voltage Across Each Element
Let current: $I = I_0\sin(\omega t)$
Across R: $V_R = IR$ (in phase with I)
$$V_R = I_0R\sin(\omega t)$$Across L: $V_L = IX_L$ (leads I by 90°)
$$V_L = I_0X_L\sin(\omega t + 90°) = I_0X_L\cos(\omega t)$$Across C: $V_C = IX_C$ (lags I by 90°)
$$V_C = I_0X_C\sin(\omega t - 90°) = -I_0X_C\cos(\omega t)$$Phasor Addition
Cannot simply add $V_R + V_L + V_C$ because phases differ!
Phasor diagram:
- $V_R$ along current direction (reference)
- $V_L$ perpendicular upward (leads by 90°)
- $V_C$ perpendicular downward (lags by 90°)
Net reactive voltage: $V_L - V_C$ (they oppose each other)
Total voltage (by Pythagorean theorem):
$$V_0 = \sqrt{V_R^2 + (V_L - V_C)^2}$$ $$V_0 = \sqrt{(I_0R)^2 + (I_0X_L - I_0X_C)^2}$$ $$V_0 = I_0\sqrt{R^2 + (X_L - X_C)^2}$$Impedance Z:
$$Z = \frac{V_0}{I_0} = \sqrt{R^2 + (X_L - X_C)^2}$$ $$Z = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$Impedance is the AC equivalent of resistance - it relates peak (or RMS) voltage to peak (or RMS) current.
Memory Trick: “Z is Hypotenuse”
Z = Zzzz (sleep on) right triangle
- Horizontal side: R
- Vertical side: $X_L - X_C$
- Hypotenuse: Z
Phase Angle
The applied voltage leads or lags the current by angle φ.
Three cases:
- $X_L > X_C$ (Inductive): $\phi > 0$, voltage leads current
- $X_L < X_C$ (Capacitive): $\phi < 0$, voltage lags current
- $X_L = X_C$ (Resonance): $\phi = 0$, voltage and current in phase
Voltage-Current relationship:
$$V = V_0\sin(\omega t + \phi)$$ $$I = I_0\sin(\omega t)$$Power in LCR Circuit
Instantaneous power: $P(t) = V(t) \cdot I(t)$
$$P(t) = V_0\sin(\omega t + \phi) \cdot I_0\sin(\omega t)$$Average power (over one cycle):
$$\langle P \rangle = \frac{V_0I_0}{2}\cos\phi = V_{rms}I_{rms}\cos\phi$$Power factor: $\cos\phi$
Components:
- Active power (consumed): $P = V_{rms}I_{rms}\cos\phi$ (dissipated in R)
- Reactive power (not consumed): $Q = V_{rms}I_{rms}\sin\phi$ (oscillates in L and C)
Note: Only resistor consumes power! L and C store and return energy.
Power Factor
$$\cos\phi = \frac{R}{Z} = \frac{R}{\sqrt{R^2 + (X_L - X_C)^2}}$$Special cases:
- $\cos\phi = 1$: Resonance (maximum power transfer)
- $\cos\phi = 0$: Pure reactive (no power consumption)
Importance: Power companies want high power factor (close to 1) for efficiency.
Resonance in Series LCR Circuit
Resonance condition: $X_L = X_C$
$$\omega_r L = \frac{1}{\omega_r C}$$ $$\omega_r^2 = \frac{1}{LC}$$Resonant angular frequency:
$$\omega_r = \frac{1}{\sqrt{LC}}$$Resonant frequency:
$$f_r = \frac{1}{2\pi\sqrt{LC}}$$At resonance:
- Impedance minimum: $Z = R$ (purely resistive)
- Current maximum: $I_{max} = \frac{V_0}{R}$
- Phase angle zero: $\phi = 0$ (V and I in phase)
- Power factor unity: $\cos\phi = 1$
- Power maximum: $P = \frac{V_{rms}^2}{R}$
Physical interpretation:
- At resonance, energy oscillates between L and C
- Voltages $V_L$ and $V_C$ are equal and opposite (can be much larger than source voltage!)
- All source voltage drops across R
Voltage Magnification
At resonance: $V_L = V_C = I_{max}X_L = \frac{V_0}{R} \cdot \omega_r L$
$$V_L = V_C = V_0 \cdot \frac{\omega_r L}{R} = V_0 \cdot Q$$where $Q = \frac{\omega_r L}{R}$ is the quality factor.
Voltage amplification: $V_L$ and $V_C$ can be Q times larger than source voltage!
Quality Factor (Q-factor)
Physical meaning:
- Measure of sharpness of resonance
- Ratio of energy stored to energy dissipated per cycle
- Higher Q → sharper peak, better frequency selectivity
Bandwidth: $\Delta\omega = \frac{\omega_r}{Q}$
Narrow bandwidth (high Q): Highly selective (radio tuning) Wide bandwidth (low Q): Less selective (broad response)
Resonance Curve
Current vs Frequency:
I
^
| Peak at f_r
| /\
| / \
| / \
| / \
|------/--------\-------> f
f_r
Characteristics:
- Peak at $f = f_r$
- High Q → Sharp, tall peak
- Low Q → Broad, flat peak
Series vs Parallel LCR Circuits
Parallel LCR Circuit
Same voltage across all elements.
Currents add (phasor addition):
$$I_0 = \sqrt{I_R^2 + (I_C - I_L)^2}$$Admittance (Y = 1/Z):
$$Y = \sqrt{\frac{1}{R^2} + \left(\omega C - \frac{1}{\omega L}\right)^2}$$At resonance:
- $I_L = I_C$ (reactive currents cancel)
- Impedance maximum (opposite of series!)
- Current minimum from source
Comparison Table
| Property | Series LCR | Parallel LCR |
|---|---|---|
| Same quantity | Current | Voltage |
| Resonance condition | $X_L = X_C$ | $X_L = X_C$ |
| Resonant frequency | $f_r = \frac{1}{2\pi\sqrt{LC}}$ | $f_r = \frac{1}{2\pi\sqrt{LC}}$ |
| At resonance | Z minimum, I maximum | Z maximum, I minimum |
| Impedance at resonance | Z = R (min) | Z = ∞ (ideally, R max practically) |
| Application | Acceptor circuit (passes $f_r$) | Rejector circuit (blocks $f_r$) |
Common Mistakes Students Make
❌ Mistake 1: Adding $V_R$, $V_L$, $V_C$ algebraically ✅ Reality: Must use phasor (vector) addition
❌ Mistake 2: Thinking impedance = $R + X_L + X_C$ ✅ Reality: $Z = \sqrt{R^2 + (X_L - X_C)^2}$
❌ Mistake 3: Confusing series and parallel resonance ✅ Reality:
- Series: Z minimum, I maximum
- Parallel: Z maximum, I minimum
❌ Mistake 4: Using wrong formula for power ✅ Reality: $P = V_{rms}I_{rms}\cos\phi = I_{rms}^2R$ (NOT $I_{rms}^2Z$)
❌ Mistake 5: Forgetting phase angle sign ✅ Reality:
- Inductive ($X_L > X_C$): $\phi > 0$ (V leads I)
- Capacitive ($X_L < X_C$): $\phi < 0$ (V lags I)
Practice Problems
Level 1: JEE Main (Basics)
Problem 1.1: A series LCR circuit has R = 10 Ω, L = 0.1 H, C = 100 μF. Find the resonant frequency.
Solution
Given: $R = 10$ Ω, $L = 0.1$ H, $C = 100$ μF $= 100 \times 10^{-6}$ F
Resonant frequency:
$$f_r = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.1 \times 100 \times 10^{-6}}}$$ $$f_r = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{1}{2\pi \times 10^{-2.5}}$$ $$f_r = \frac{1}{2\pi \times 3.16 \times 10^{-3}} = \frac{1000}{2\pi \times 3.16}$$ $$f_r = \frac{1000}{19.87} \approx 50.3 \text{ Hz}$$Answer: 50.3 Hz
Problem 1.2: In a series LCR circuit, R = 30 Ω, $X_L = 50$ Ω, $X_C = 30$ Ω. Find: (a) Impedance (b) Phase angle
Solution
Given: $R = 30$ Ω, $X_L = 50$ Ω, $X_C = 30$ Ω
(a) Impedance:
$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{30^2 + (50 - 30)^2}$$ $$Z = \sqrt{900 + 400} = \sqrt{1300} = 36.06 \text{ Ω}$$(b) Phase angle:
$$\tan\phi = \frac{X_L - X_C}{R} = \frac{50 - 30}{30} = \frac{20}{30} = \frac{2}{3}$$ $$\phi = \tan^{-1}(2/3) = 33.69°$$Since $X_L > X_C$, circuit is inductive, so voltage leads current by 33.69°.
Answers: (a) 36.06 Ω, (b) 33.69° (inductive)
Level 2: JEE Advanced (Application)
Problem 2.1: A series LCR circuit with R = 20 Ω, L = 0.2 H, C = 50 μF is connected to a 230 V, 50 Hz AC source. Calculate: (a) Inductive and capacitive reactances (b) Impedance (c) Current (d) Power consumed
Solution
Given: $R = 20$ Ω, $L = 0.2$ H, $C = 50 \times 10^{-6}$ F, $V_{rms} = 230$ V, $f = 50$ Hz
$\omega = 2\pi f = 100\pi$ rad/s
(a) Reactances:
$$X_L = \omega L = 100\pi \times 0.2 = 20\pi = 62.83 \text{ Ω}$$ $$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 50 \times 10^{-6}} = \frac{10^4}{5\pi} = \frac{2000}{\pi} = 63.66 \text{ Ω}$$(b) Impedance:
$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (62.83 - 63.66)^2}$$ $$Z = \sqrt{400 + (-0.83)^2} = \sqrt{400 + 0.69} \approx 20.02 \text{ Ω}$$Almost at resonance! ($X_L \approx X_C$)
(c) Current:
$$I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{20.02} = 11.49 \text{ A}$$(d) Power:
$$P = I_{rms}^2 R = (11.49)^2 \times 20 = 132 \times 20 = 2640 \text{ W}$$Or: $\cos\phi = \frac{R}{Z} = \frac{20}{20.02} \approx 0.999$
$$P = V_{rms}I_{rms}\cos\phi = 230 \times 11.49 \times 0.999 = 2640 \text{ W}$$✓
Answers: (a) 62.83 Ω, 63.66 Ω, (b) 20.02 Ω, (c) 11.49 A, (d) 2640 W
Problem 2.2: At resonance in a series LCR circuit, the voltage across L is 200 V and across C is 200 V. If R = 10 Ω and the current is 2 A, find the applied voltage.
Solution
Given: At resonance, $V_L = 200$ V, $V_C = 200$ V, $R = 10$ Ω, $I = 2$ A
At resonance:
- $V_L = V_C$ ✓ (confirmed)
- They cancel each other in phasor addition
- Impedance: $Z = R = 10$ Ω
Applied voltage:
$$V = IR = 2 \times 10 = 20 \text{ V}$$Key insight: Even though $V_L = V_C = 200$ V (10 times the source!), they are 180° out of phase and cancel each other. Only voltage across R matters.
Answer: 20 V
Voltage magnification: $Q = \frac{V_L}{V} = \frac{200}{20} = 10$
Level 3: JEE Advanced (Challenging)
Problem 3.1: A series LCR circuit has L = 1 H, C = 1 μF, R = 100 Ω. Find: (a) Resonant frequency (b) Quality factor (c) Bandwidth (d) If connected to 10 V AC at resonance, find voltage across L
Solution
Given: $L = 1$ H, $C = 1 \times 10^{-6}$ F, $R = 100$ Ω, $V = 10$ V
(a) Resonant frequency:
$$\omega_r = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1 \times 10^{-6}}} = \frac{1}{10^{-3}} = 1000 \text{ rad/s}$$ $$f_r = \frac{\omega_r}{2\pi} = \frac{1000}{2\pi} = 159.15 \text{ Hz}$$(b) Quality factor:
$$Q = \frac{\omega_r L}{R} = \frac{1000 \times 1}{100} = 10$$(c) Bandwidth:
$$\Delta\omega = \frac{\omega_r}{Q} = \frac{1000}{10} = 100 \text{ rad/s}$$ $$\Delta f = \frac{\Delta\omega}{2\pi} = \frac{100}{2\pi} = 15.92 \text{ Hz}$$(d) Voltage across L: At resonance: $I = \frac{V}{R} = \frac{10}{100} = 0.1$ A
$$V_L = IX_L = I\omega_r L = 0.1 \times 1000 \times 1 = 100 \text{ V}$$Or: $V_L = QV = 10 \times 10 = 100$ V ✓
Voltage magnification by factor of 10!
Answers: (a) 159.15 Hz, (b) 10, (c) 15.92 Hz, (d) 100 V
Problem 3.2: Show that at resonance in a series LCR circuit, the power dissipated equals $\frac{V_{rms}^2}{R}$.
Solution
At resonance:
- Condition: $X_L = X_C$
- Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + 0} = R$
- Phase angle: $\tan\phi = \frac{X_L - X_C}{R} = 0$ → $\phi = 0°$
- Power factor: $\cos\phi = \cos(0°) = 1$
Current:
$$I_{rms} = \frac{V_{rms}}{Z} = \frac{V_{rms}}{R}$$Power dissipated:
$$P = V_{rms}I_{rms}\cos\phi = V_{rms} \cdot \frac{V_{rms}}{R} \cdot 1 = \frac{V_{rms}^2}{R}$$Alternative:
$$P = I_{rms}^2 R = \left(\frac{V_{rms}}{R}\right)^2 R = \frac{V_{rms}^2}{R}$$✓
Conclusion: At resonance, LCR circuit behaves like pure resistor, dissipating maximum power.
Applications of LCR Circuits
- Radio Tuning: Adjust C to match broadcast frequency
- Filters:
- Band-pass: Pass frequencies near resonance
- Band-stop: Block frequencies near resonance
- Oscillators: Generate specific frequencies
- Impedance Matching: Maximize power transfer
- Signal Processing: Separate frequency components
- Communication Systems: Modulation and demodulation
- Medical Devices: MRI, ultrasound imaging
- Wireless Charging: Resonant inductive coupling
Real-World Example: Radio Receiver
How AM radio works:
- Antenna receives all radio frequencies
- LCR tuning circuit (variable capacitor):
- Adjust C to set resonant frequency = desired station
- At resonance, that frequency has maximum current
- Detector extracts audio signal
- Amplifier boosts signal
- Speaker converts to sound
Why it works: High Q-factor makes circuit highly selective, rejecting unwanted frequencies.
Connection to Other Chapters
- AC Circuits: Foundation for L and C behavior
- Self Inductance: Inductive element properties
- Capacitors: Capacitive element properties
- Oscillations: Resonance is fundamental oscillation phenomenon
- Electromagnetic Waves: LC oscillations generate EM waves
Key Takeaways
✓ Series LCR: $Z = \sqrt{R^2 + (X_L - X_C)^2}$ ✓ Phase angle: $\tan\phi = \frac{X_L - X_C}{R}$ ✓ Power: $P = V_{rms}I_{rms}\cos\phi = I_{rms}^2R$ (only R dissipates) ✓ Resonance: $f_r = \frac{1}{2\pi\sqrt{LC}}$ (when $X_L = X_C$) ✓ At resonance: Z minimum, I maximum, $\phi = 0°$, power maximum ✓ Quality factor: $Q = \frac{\omega_r L}{R}$ (sharpness of resonance) ✓ Voltage magnification: $V_L = V_C = QV$ at resonance ✓ Series: Current resonance (I max); Parallel: Voltage resonance (Z max)
Formula Summary
Series LCR Circuit:
Impedance:
$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2}$$Phase angle:
$$\tan\phi = \frac{X_L - X_C}{R}$$Power:
$$P = V_{rms}I_{rms}\cos\phi = I_{rms}^2R$$Power factor:
$$\cos\phi = \frac{R}{Z}$$Resonance:
$$f_r = \frac{1}{2\pi\sqrt{LC}}, \quad \omega_r = \frac{1}{\sqrt{LC}}$$At resonance:
- $Z = R$ (minimum)
- $I = \frac{V}{R}$ (maximum)
- $\phi = 0°$
- $\cos\phi = 1$
Quality factor:
$$Q = \frac{\omega_r L}{R} = \frac{1}{\omega_r CR} = \frac{1}{R}\sqrt{\frac{L}{C}}$$Bandwidth:
$$\Delta\omega = \frac{\omega_r}{Q}$$Next Topic: Transformers - Voltage transformation, efficiency, and power transmission principles